Problems & Puzzles: Conjectures

Conjecture 107. The  Shyam's conjecture about square-free triangular numbers

On Nov 29, 2024 Shyam Sunder Gupta sent the following conjecture:


The nth triangular number, represented by Tn, equals n*(n+1)/2.
Consider a set of three consecutive triangular numbers Tn, T(n+1), and T(n+2), such that
the number of prime factors (counting multiplicities) of each of the three consecutive
triangular numbers is three. The smallest set of such three consecutive triangular
numbers is as follows:
T17 = 153 = 3 * 3 * 17, T18 = 171 = 3 * 3 * 19, and T19 = 190 = 2 * 5 * 19.
It can be seen that T19 is square-free, whereas T17 and T18 are not square-free.
The next set of three consecutive triangular numbers, each with three prime factors
(counting multiplicities), is (406, 435, 465) as
T28 = 406 = 2 * 7 * 29, T29 = 435 = 3 * 5 * 29, and T30 = 465 = 3 * 5 * 31.
It can be seen that T28, T29, and T30 are all square-free.
Reference can be made to OEIS, where I submitted the sequence (A349696) of 
"Smallest number in a set of three consecutive triangular numbers, each with three prime
factors (counted with multiplicity)." with a comment that 153 is the only known number in
the sequence that is not square-free.

Conjecture:

"For a set of three consecutive triangular numbers Tn, T(n+1), and T(n+2), such that if the number of prime factors (counting multiplicities) of each of the three consecutive triangular numbers is three, then except for the smallest set, i.e., (153, 171, 190), all three triangular numbers in all other sets are square-free."

(I have tested it for triangular numbers < 10^17, and the conjecture holds.)

1) Can this conjecture be proved or disproved?
2) In case this conjecture is disproved, can someone find a counterexample?


From 21-27 Dec,. 2024 contributions came from Fred Schneider, Alain Rochelli, Michael Branicky, Alessandro Casini, Emmanuel Vantieghem

 

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Fred wrote:

First consider the formulas of three consecutive triangular numbers.
 
T(n) = n * (n + 1) /2
 
T(n+1) = (n+ 1) * (n + 2) /2
 
T(n+1) = (n+ 2) * (n + 3) /2
 
These three triangular numbers contain the factors of n through n + 3 (barring a single 2 in the denominator).  Note that among 4 consecutive numbers, one has to be a multiple of 4, two have to be even, and at least one has to be a multiple of 3.

The convention used below is that p, p1, p2 and p3 are all prime numbers with p being the squared factor.

The non-square free solution had this form for the four consecutive numbers:

n = p1
n+1 = 2 * p * p
n+2 = p2
n+3 = 4 * p3
 
resulting in the three consecutive triangular numbers, each with exactly three prime factors:
 
T(n) = n * (n+1) / 2 = p1 * p * p
T(n+1) = (n+1) * (n+2) / 2 = p2 * p * p
T(n+2) = (n+2) * (n+3) / 2 =  2 * p2 * p3
 
where n = 17, p1 = 17, p2 = 19 , p = 3 and p3 = 5
 
A counterexample to this conjecture would either have to have the above arrangement for the four consecutive numbers or one of these other three arrangements.
 
(Arrangement #2)
 
n = p1
n+1 = 4 * p2
n+2 = p3
n+3 = 2 * p * p
 
(Arrangement #3)
 
n = 2 * p * p
n+1 = p1
n+2 = 4 * p2
n+ 3 = p3
 
(Arrangement #4)
 
n = 4 * p1
n+1 = p2
n+2 = 2 * p * p
n+ 3 = p3
 
As p is a squared prime factor, it must be always less than p1, p2 and p3 in one of these arrangements beyond a trivial threshold.  (At n= 6, p = 2 and we get non-prime p1 = 1.5.  At n = 7, p = 2 and p2 = 2.5)
 
As stated before, each set of four consecutive numbers must have one number divisible by 3.  So, either p is 3 or p = 2 and one of the other primes is 3.
 
If p = 2 then 2 * p * p = 8.  This is not possible because a multiple of 4 is always two away from it in each arrangement.
 
So, p must be 3. As the known solution is for the first arrangement, we check the remaining arrangements:
 
2) n + 3 = 2 * p * p = 18, n = 15 (not a prime)
 
3) n = 2 * p * p = 18, n + 3 = 21 (not a prime)

4) n + 2 = 2 * p * p = 18, n = 16 which is of the form 4 * x. but x = 4 is not a prime

So, there are no possible counterexamples and the conjecture is true.

***

Alain wrote:

This conjecture is stated in A349696 and completed in A255200. Furthermore, we have T(n)= A000217 ( A255200 (n)).

It is easy to prove that Tn is always divisible by 2 or 3.

The generic notation of T(n) is then T(n)=2*q*r or T(n)=3*p*r with 3<p<q<r if the conjecture is true. If it is false we have p=r or q=r.

First case: let T(n)=n*(n+1)/2=2*r^2

--> n*(n+1)=(2*r)^2  which is not possible with n integer and r prime (i.e. n*(n+1) is never a square).

Second case: let T(n)=n*(n+1)/2=3*r^2

--> T(n+1)=(n+1)*(n+2)/2 = [n*(n+1)/2]*[(n+2)/n] = T(n)*(n+2)/n = 3*r^2*(n+2)/n

T(n+1) is 3-almost prime only if n=r with r and r+2 twin primes --> T(n+1)=3*r*(r+2)

T(n+2)=(n+2)*(n+3)/2  = [(n+1)*(n+2)/2]*[(n+3)/(n+1)] = T(n+1)*(n+3)/(n+1) = 3*r*(r+2)*(n+3)/(n+1)

With n=r --> T(n+2) = 3*r*(r+2)*(r+3)/(r+1) which is never a 3-almost prime.

In conclusion, it seems that this conjecture is true so it is no longer a conjecture.

 

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Michael wrote:

I tested the conjecture up to T(867608681) = 376372412107084221 > 3.76 * 10^17.
This took only 24 min using the 10000-term b-file at https://oeis.org/A255200.
A search from scratch that exceeds that limit would take >24 hr on the same hardware.

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Alessandro wrote:

The conjecture is true. Between (n, n+1, n+2, n+3), i.e. the numbers that in pairs form the triangular numbers, one must be divisible by 4 (contributing a factor of 2 in triangular ones) e one (two in some cases) is a multiple of 3. With these constraints, it's easy to see that, except for a finite number of cases, the only ways of constructing triangular numbers are (n, n+1, n+2, n+3) of the form (p, 6q, s, 4t) or (4p, q, 6s, t), with p, q, s, t primes. Therefore, the only way to have a no square-free triangular numbers is among the finite cases, in which only n=17 appears.

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Emmanuel wrote:

The conjecture is true !
We shall prove that there exist only two consecutive triangular numbers that have three prime factors
and that are non square-free.
 
Proof.
Let's examine how a triangular number  t   can be a non square-free product of three primes.

1) t = p^3
    This is totally impossible since  t  has to be of the form  n(n+1)/2 :
    Such numbers must have at least two different prime factors.

2) t = 4p, with  p  an odd prime.
    Since  1 + 8t = 1 + 32p  must be a square, say : x^2  we may deduce :
        32p = (x - 1)(x + 1).
    The only common factor of  x - 1  and  x + 1  is  2.
    So, one of them must be  2p  and the other must be  16.
    This can only be true if  p = 7  or : t =28 = 7*8/2.
    In that case, the former triangular is  21 (product of two primes)
    and the next triangular is  36 (four primes)

3) t = 2 p^2, with  p  an odd prime.
    Then, 1 + 16 p^2  must be of the form  x^2  and thus :
        16p^2 = (x - 1)(x + 1).
    The only possibility to achieve that is  that one of the two numbers  x - 1, x + 1  must be  2 p^2
    while the other must be  8, totally impossible !

4) T = p q^2  with  p, q  odd primes.
    Thus, we must have : 8 p q^2 = (x - 1)(x + 1).  There are four possibilities :
       a) 4p = x + 1, 2q^2 = x - 1
           or : 4q^2 = 4p - 2
           or : 2q^2 = 2p - 1,  completely impossible      
       b) 2p = x + 1, 4q^2 = x - 1
           or : 4q^2 = 2p - 2
           or : 2q^2 + 1 = p.  Only  p = 19, q = 3   seems to satisfy this
           (look at the equation modulo 3)
       c) 4p = x - 1, 2q^2 = x + 1
           or : 4q^2 = 4p + 2
           or : 2q^2 = 2p + 1 !  Totally impossible
       d) 2p = x - 1, 4q^2 = x + 1
           or : 4q^2 = 2p + 2
           or : 2q^2 - 1 = p.  There are many many solutions : see  https://oeis.org/A092057.
           It is easy to show that, in every case  t  will be  (2q^2 - 1) q^2.
           Thus, the next triangular then will be  q^2 (2q^2 + 1).  And this is a product of three primes
           if and only if  (2q^2 + 1)  is a prime : i. e. : p  and  p + 2 must be twin primes.
           And this is only possible when  q = 3.
           
   Summarizing, we have proved that, if  t, t'  are two consecutive triangular numbers with only three prime factors,
   then  t = 17*9  and  t = 9*19, Q. E. D.

 

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